THE SANTALO-REGIONS OF A CONVEX BODY · 2018. 11. 16. · Abstract. Motivated by the Blaschke-Santal o inequality, we de ne for a con-vex body K in Rn and for t 2 R the Santal o-regions

TRANSACTIONS OF THEAMERICAN MATHEMATICAL SOCIETYVolume 350, Number 11, November 1998, Pages 4569–4591S 0002-9947(98)02162-X

THE SANTALÓ-REGIONS OF A CONVEX BODY

MATHIEU MEYER AND ELISABETH WERNER

Abstract. Motivated by the Blaschke-Santaló inequality, we define for a con-vex body K in Rn and for t ∈ R the Santaló-regions S(K, t) of K. We in-vestigate the properties of these sets and relate them to a concept of affinedifferential geometry, the affine surface area of K.

Let K be a convex body in Rn. For x ∈ int(K), the interior of K, let Kx bethe polar body of K with respect to x. It is well known that there exists a uniquex0 ∈ int(K) such that the product of the volumes |K||Kx0 | is minimal (see forinstance [Sch]). This unique x0 is called the Santaló-point of K.

Moreover, the Blaschke-Santaló inequality says that |K||Kx0| ≤ v2n (where vndenotes the volume of the n-dimensional Euclidean unit ball B(0, 1)) with equalityif and only if K is an ellipsoid.

For t ∈ R we consider here the sets

S(K, t) = {x ∈ K : |K||Kx|

v2n≤ t}.

Following E. Lutwak, we call S(K, t) a Santaló-region of K.Observe that it follows from the Blaschke-Santaló inequality that the Santaló-

point x0 ∈ S(K, 1), and that S(K, 1) = {x0} if and only if K is an ellipsoid. ThusS(K, t) has non-empty interior for some t < 1 if and only if K is not an ellipsoid.

In the first part of this paper we show some properties of S(K, t) and giveestimates on the “size” of S(K, t). This question was asked by E. Lutwak.

In the second part we show how S(K, t) is related to the affine surface area ofK.

The affine surface area as(K) is originally a notion of differential geometry. Fora convex body K in Rn with sufficiently smooth boundary ∂K it is defined as

as(K) =∫∂K

κ(x)1

n+1dµ(x),

where κ(x) is the Gaussian curvature at x ∈ ∂K and µ is the surface measure on ∂K.The affine surface area is invariant under affine transformations with determinant1. It arises naturally in questions concerning the approximation of convex bodiesby polytopes (see [G]) and in a priori estimates of PDE’s ([Lu-O]).

Received by the editors October 25, 1996.1991 Mathematics Subject Classification. Primary 52A20; Secondary 52A38.Key words and phrases. Blaschke-Santaló inequality, affine surface area.Supported by a grant from the National Science Foundation.The paper was written while both authors were at MSRI.

c©1998 American Mathematical Society

4569

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use

4570 MATHIEU MEYER AND ELISABETH WERNER

It has been one of the aims of convexity theory to extend the notions of differ-ential geometry and (for instance) of affine surface area to arbitrary convex bodieswithout any smoothness assumptions on the boundary.

Within the last few years four different extensions have been given (due toLeichtweiss [L1], Lutwak [Lu], Schütt-Werner [S-W] and Werner [W]), and it wasshown that they all coincide ([S1], [D-H]).

We give here another such extension, arising again from a completely differentcontext. It will also follow that this new extension coincides with the others.

The authors wish to thank MSRI for its hospitality, and the organizers of thespecial semester in Convex Geometry and Geometric Functional Analysis at MSRIfor inviting them. It was during our stay there that the paper was written.

Unless stated otherwise we will always assume that a convex body K in Rn hasits Santaló-point at the origin. Then 0 is the center of mass of the polar body K0,which may be written as∫

K0〈x, y〉dy = 0 for every x ∈ Rn.

By |K| we denote the n-dimensional volume of K. hK is the support function ofK. If K is centrally symmetric, ||.||K is the norm on Rn that has K as its unit ball.By ||.|| we denote the standard Euclidean norm on Rn, and 〈.,. 〉 is the usual innerproduct on Rn. B(a, r) is the n-dimensional Euclidean ball with radius r centeredat a. For x ∈ K, Kx = (K − x)0 = {y ∈ Rn : 〈y, z − x〉 ≤ 1for all z ∈ K} is thepolar body of K with respect to x; K0 denotes the polar body with respect to theSantaló-point. Moreover for u ∈ Sn−1 we will denote by φuK(y) or for short by φ(y)the (n− 1)-dimensional volume of the sections of K orthogonal to u; that is,

φ(y) = φuK(y) = |{z ∈ K : 〈z, u〉 = y}|.

I. Properties of the Santaló-regions

We start by listing some of the properties of S(K, t). Recall that for δ > 0, δsmall enough, Kδ is said to be a (convex) floating body of K, if it is the intersectionof all halfspaces whose defining hyperplanes cut off a set of volume δ of K ([S-W]).More precisely, for u ∈ Sn−1 and for 0 < δ let auδ be defined by

|{x ∈ K : 〈x, u〉 ≥ auδ }| = δ.Then Kδ =

⋂u∈Sn−1{x ∈ K : 〈x, u〉 ≤ auδ }.

In the following proposition we consider only those t ∈ R for which S(K, t) 6= ∅.Proposition 1. Let K be a convex body in Rn. Then:

(i) S(K, t) is strictly convex for all t.(ii) S(A(K), t) = A(S(K, t)) for all regular affine transformations A, for all t.(iii) The boundary of S(K, t) is C∞ for all t.(iv) t 7−→ S(K, t) is increasing and concave; that is, for all t, s and for all α ∈ R,

0 ≤ α ≤ 1,S(K,αt+ (1 − α)s) ⊃ αS(K, t) + (1− α)S(K, s).


THE SANTALÓ-REGIONS OF A CONVEX BODY 4571

(v) For all 0 < δ < 12 , Kδ|K| is contained in S(K,1

4δ(1−δ) ).

Proof. We frequently use the following well known formula. For all x ∈ int(K)

|Kx| = 1n

∫Sn−1

dσ(u)(hK(u)− 〈u, x〉)n ,(1)

where σ is the spherical Lebesgue measure.Indeed, |Kx| = 1n

∫Sn−1

dσ(u)

(h̃K(u))n, where h̃K is the support function of K centered

at x. Now observe that h̃K(u) = hK(u)− 〈u, x〉; thus (1) follows.(i) Observe that for all u ∈ Sn−1 the function

x 7−→ 1(hK(u)− 〈x, u〉)n

is convex on int(K). (1) then implies that

x 7−→ |Kx| = 1n

∫Sn−1

dσ(u)(hK(u)− 〈u, x〉)n

is convex. In fact the function

x 7−→ |Kx| = 1n

∫Sn−1

dσ(u)(hK(u)− 〈u, x〉)n

is strictly convex, as for x, y ∈ int(K), x 6= y,σ({u ∈ Sn−1 : 〈u, x〉 = 〈u, y〉}) = 0.

Therefore (i) follows.(ii) Let A be a one-to-one affine transformation. We can write A = L+ a, where

L is a one-to-one linear transformation and a is a vector in Rn. Then

(A(K))y = {x ∈ Rn : 〈x,Az − y〉 ≤ 1 for all z ∈ K}= {x ∈ Rn : 〈L∗x, z −A−1y〉 ≤ 1 for all z ∈ K}

= (L∗)−1(KA−1(y)).

Hence

S(AK, t) = {y ∈ AK : |AK||AKy|

v2n≤ t}

= {y ∈ AK : |K||KA−1(y)|v2n

≤ t}= A(S(K, t)).

(iii) Let K, n and t be fixed. By (1), ∂S(K, t) = {x ∈ K : F (x) = ntv2n|K| }, whereF (x) =

∫Sn−1

dσ(u)(hK(u)−〈u,x〉)n . F is continuous on int(K) with continuous partial

derivatives of all orders, has a unique minimum at the Santaló-point x0, and isconvex (see (i)). Therefore (iii) follows from the implicit function theorem.

(iv) is obvious from the proof of (i).(v) Let δ ∈ (0, 12 ), and let H be a hyperplane that has non-empty intersection

with K and is such that|K ∩H+| = δ|K|,

where H+ is one of the two halfspaces determined by H . By definition the convexfloating body Kδ|K| is the intersection of all the halfspaces H− determined by allsuch hyperplanes H .



On the other hand, by [Me-P] there exists x ∈ H ∩ int(K) such that|K||Kx|v2n

≤ 14δ(1− δ) .

This means that x ∈ S(K, 14δ(1−δ) ). Consequently

Kδ|K| ⊆ S(K,1

4δ(1− δ) ).

Remark 2. (i) Proposition 1,(iv) says that Kδ|K| ⊆ S(K, 14δ(1−δ ). We will show(see Proposition 14) that in the case of the convex body with sufficiently smoothboundary and positive Gaussian curvature everywhere a converse inclusion holdsfor δ “small”.

(ii) Note also that for K = B(0, 1),

(B(0, 1))δvn ∼ S(B(0, 1),vn−1

δ(n+ 1)vn),

for δ sufficiently small. More precisely, for δ ≤ 2n+1vn−1√e(n+1)vnn

n+12

S(B(0, 1),vn−1√

eδ(n+ 1)vn) ⊆ (B(0, 1))δvn ⊆ S(B(0, 1),

(√e)

n+1n−1 vn−1

δ(n+ 1)vn).

This follows from the forthcoming Corollary 5 and from the fact that the volumeof a cap of the Euclidean unit ball of height ∆ can be estimated from above by

vn−1n+ 1

(2∆)n+1

2

and from below byvn−1n+ 1

(2∆)n+1

2 (1− ∆2

)n−1

2 .

For δ “close” to 12 ,

(B(0, 1))δvn ∼ S(B(0, 1),1

4δ(1− δ) ).

More precisely, let ε ≤ 1√n

and 12 > δ ≥ 12 − ε vn−1√evn . Then

(B(0, 1))δvn ⊆ S(B(0, 1),1

4δ(1− δ) ) ⊆8vn−1√n+ 1vn

(B(0, 1))δvn .

The following lemmas will enable us to compute |(B(0, 1))x|. They are alsoneeded for Part II.

Lemma 3. Let x ∈ int(K). Then

|Kx| =∫K0

dy

(1 − 〈x, y〉)n+1 .

Proof. By (1)



|Kx| = 1n

∫Sn−1

dσ(u)(hK(u))n(1− 〈 uhK(u) , x〉)n

=∫Sn−1

∫ 1hK (u)

0

rn−1

(1− 〈x, ru〉)n+1 drdσ(u) =∫K0

dy

(1− 〈x, y〉)n+1 .

Remark. We will use Lemma 3 mostly in the following form:Let u ∈ Sn−1 and λ ∈ R be such that x = λu ∈ int(K). Then Lemma 3 says

that

|Kx| =∫ hK0 (u)−hK0 (−u)

φuK0(t)dt(1− λt)n+1 ,(2)

where φuK0(t) = |{z ∈ K0 : 〈z, u〉 = t}|.Lemma 4. (i) Let 0 ≤ α < 1. Then∫ 1

−1

(1− x2)n−12 dx(1− αx)n+1 =

2n(Γ(n+12 ))2

(1− α2)n+12 n!(ii) For α ∈ (0, 1) let

I(α) = (∫ 1

0

(1− x2)n−12 dx(1− αx)n+1 )(

αn+1

2 (1− α)n+12 n!2

n−12 (Γ(n+12 ))

2).

ThenI(α) ≤ 1 and lim

α→1I(α) = 1.

(iii) Let a, b > 0, n ∈ N and λa < 1. Then∫ a−b

(1− ya )n−1(1− λy)n+1 dy =

(a+ b)n

nan−1(1− λa)(1 + λb)n .

Proof. (i) We put x = (1 − 1−α1+αu)/(1 + 1α1+αu). This gives (i).(ii) Put x = 1− w 1−αα . Then

I(α) =n!

2n−1

2 (Γ(n+12 ))2

∫ α1−α

0

wn−1

2 (2 − 1−αα w)n−1

2 dw

(1 + w)n+1.

The upper estimate for (ii) follows immediately from this last expression. And bythe monotone convergence theorem this last expression tends to

n!(Γ(n+12 ))

2

∫ ∞0

wn−1

2 dw

(1 + w)n+1,

which is equal to 1.(iii) Note that ∫

(1− ya )n−1(1− λy)n+1 dy =

(1− ya )nn(λ− 1a )(1− λy)n

.

This immediately implies (iii).



Corollary 5. Let B(0, r) be the n-dimensional Euclidean ball with radius r centeredat 0. For u ∈ Sn−1 let x = λu, 0 ≤ λ < r. Then

|(B(0, r))x| = vnrn(1 − (λr )2)

n+12

.

Proof. The proof follows from (2) and Lemma 4 (i).

Next we estimate the “size” of S(K, t) in terms of ellipsoids. Recall that for aconvex body K the Binet ellipsoid E(K) is defined by (see [Mi-P])

||u||2E(K) =1|K|

∫K

〈x, u〉2dx, for all u ∈ Rn.We first treat the case when K is a symmetric convex body.

Theorem 6. Let K be a symmetric convex body in Rn. For all t ∈ Rdn(t)E(K0) ⊆ S(K, t) ⊆ cn(t)E(K0),

where

dn(t) =1√3n

(1− |K||K0|

tv2n)

12

and

cn(t) = min{( 2(n+ 1)(n+ 2))12 (

tv2n|K||K0| )

12 (1− |K||K

0|tv2n

)12 ,√

2(1− ( |K||K0|

tv2n)

1n )

12 }.

Remarks. (i) In particular, for any ellipsoid E, S(E, 1) = {0}.(ii) If t→ |K||K0|v2n , then S(K, t) → {0}.(iii) The second expression in cn(t) gives a better estimate from above than the

first iff |K||K0|/tv2n is of a smaller order of magnitude than (n logn)−1.(iv) Recall that for two isomorphic Banach spaces E and F the Banach-Mazur

distance d(E,F ) is defined by

d(E,F ) = inf{||T || ||T−1|| : T is an isomorphism from E onto F}.For symmetric convex bodies K, L in Rn we define

d(K,L) = d((Rn, ||.||K), (Rn, ||.||L)).Then it follows from Theorem 6 that

d(S(K, t), E(K0)) ≤ ( 6tv2n

|K||K0| )12 .

Thus for ρ ∈ R, ρ > 1,d({x ∈ K : |Kx| ≤ ρ|K0|}, E(K0)) ≤ (6ρ) 12 ,

independent of K and n. It follows that for fixed ρ, {x; |Kx| ≤ ρ|K0|} is almost anellipsoid.

Proof of Theorem 6. Let u ∈ Sn−1, λ ∈ R, 0 ≤ λ < 1||u||K and x = λu. By (2) andsymmetry,

|Kx| =∫ ||u||K

0

φ(y)(1

(1 − λy)n+1 +1

(1 + λy)n+1)dy,

where φ = φuK0 . For fixed λ ≥ 0 put

fλ(y) =1

(1 − λy)n+1 +1

(1 + λy)n+1.



Observe that fλ is increasing in y, if y ≥ 0. Put

a =n

∫∞0 φ(y)dyφ(0)

=n|K0|2φ(0)

.

Now we distinguish two cases.1. λa < 1. Then we claim that for all functions ψ : R+0 → R+0 such that ψ

1n−1

is continuous on its support and continuous from the right at 0, decreasing andconcave on its support and such that

(i) ψ(0) = φ(0),

(ii)∫ ∞

0

ψ(y)dy =∫ ∞

0

φ(y)dy =|K0|

2,

the integral∫∞0 ψ(y)fλ(y)dy is maximal if ψ is of the form

ψ0(y) ={φ(0)(1 − ya )n−1 if y ∈ [0, a],0 otherwise.

Indeed, let ψ be a function with above properties and with support on [0,ã]. Put

H(t) =∫ at

ψ(y)dy −∫ at

ψ0(y)dy.

Note that ã ≤ a, H(0) = 0 = H(a) and that the derivative of H with respect to tis first negative, then positive; therefore H(t) ≤ 0.

Consequently (with gλ(y) = fλ(y)− 2)∫ ∞0

ψ(y)gλ(y)dy =∫ ∞y=0

ψ(y)(∫ yt=0

g′λ(t)dt)dy =

∫ ∞t=0

g′λ(t)(

∫ ∞t

ψ(y)dy) dt

≤∫ ∞

0

g′λ(t)(

∫ ∞t

ψ0(y)dy) dt =∫ ∞

0

ψ0(y)gλ(y)dy.

From this the above claim follows.Hence

|Kx| ≤ φ(0)∫ a

0

((1− ya )n−1(1− λy)n+1 +

(1− ya )n−1(1 + λy)n+1

)dy =|K0|

1− λ2a2 .

Here we have used Lemma 4 (iii).For x = λu ∈ ∂S(K, t),

1 = ||x||S(K,t) = λ||u||S(K,t)(3)and

t =|K||Kx|v2n

.(4)

Therefore

t ≤ |K||K0|

v2n(1− λ2a2)and hence by (3)

||u||S(K,t) ≤ n2 (1−|K||K0|tv2n

)−12|K0|φ(0)

.

Now ([B], respectively [He]; see also [Mi-P])

|K0|φ(0)

≤ 2√

3(

∫K0

|〈x, u〉|2dx|K0| )

12 = 2

√3||u||E(K0),(5)



and thus

S(K, t) ⊇ 1√3n

(1 − |K||K0|

tv2n)

12E(K0).

2. λa ≥ 1. Again let x = λu ∈ ∂S(K, t). By definition of a, (3) and (5)||u||S(K,t) ≤

√3n||u||E(K0).

This implies that

S(K, t) ⊇ 1√3nE(K0),

which proves the inclusion from below also in this case.

On the other hand, by (1) and symmetry

|Kx| = 12n

∫Sn−1

(1

(||v||K0 − 〈v, x〉)n +1

(||v||K0 + 〈v, x〉)n )dσ(v)

≥ 12n

∫Sn−1

(2 + n(n+ 1)(〈x, v〉||v||K0 )

2)dσ(v)||v||nK0

= |K0|+ (n+ 1)(n+ 2)2

∫K0|〈x, y〉|2dy

= |K0|+ (n+ 1)(n+ 2)2

λ2|K0| ||u||2E(K0).Then (3) and (4) give

||u||S(K,t) ≥ ((n+ 1)(n+ 2))12√

2(

tv2n|K||K0| − 1)

− 12 ||u||E(K0),

or, equivalently,

S(K, t) ⊆√

2((n+ 1)(n+ 2))

12(

tv2n|K||K0| − 1)

12E(K0).

Using (2) and a minimality argument similar to the maximality argument of theabove claim, we get the other upper bound. Namely, for fixed λ and for all functionsψ : R+0 → R+0 such that ψ

1n−1 is continuous on its support and continuous from

the right at 0, decreasing and concave on its support and such that

(i) ψ(0) = φ(0),

(ii)∫ ∞

0

ψ(y)dy =∫ ∞

0

φ(y)dy =|K0|

2,

the integral∫∞0 ψ(y)fλ(y)dy is minimal if ψ is of the form

ψ(y) ={φ(0) if y ∈ [0, a],0 otherwise,

where

a =|K0|2φ(0)

.

Note that in this situation λa < 1 always.Consequently

|Kx| ≥ φ(0)∫ a

0

fλ(y)dy ≥ |K0|

(1 − λ2a2)n .



Then we again use (3), (4) and the fact that ([B], or respectively [He]; see also[Mi-P])

|K0|φ(0)

≥ √2||u||E(K0),

and get

S(K, t) ⊆√

2(1− ( |K||K0|

tv2n)

1n )

12E(K0).

Now we consider the non-symmetric case.

Theorem 7. Let K be a convex body in Rn. Then

d′n(t)E(K0) ⊆ S(K, t) ⊆ c′n(t)E(K0),

where

c′n(t) =2√

2((e− 2)(n+ 1)(n+ 2)) 12 (

tv2n|K||K0| )

12 (1− |K||K

0|tv2n

)12

and

d′n(t) = dn(t) =1√3n

(1− |K||K0|

tv2n)

12 .

Proof. By (2) we get for u ∈ Sn−1 and x = λu with 0 ≤ λ < 1/hK0(u) that

|Kx| =∫ hK0 (u)−hK0(−u)

φ(y)(1 − λy)n+1 dy,

where φ = φuK0 .Notice that K0 has its center of gravity at 0, as K has its Santaló-point at 0.

Therefore ∫ hK0(u)−hK0(−u)

yφ(y)dy = 0.

Notice also that

1 = hS(K,t)0(x) = λhS(K,t)0(u).(6)

Now we apply the following result of Fradelizi [F] to the functions φ(y) andfλ(y) = 1(1−λy)n+1 to get the same upper estimate for |Kx|as in the proof of Theo-rem 6. Therefore d′n(t) = dn(t).

Theorem ([F]). Let ψ : R → R, ψ ≥ 0, be such that ψ 1n−1 is continuous andconcave on its support and such that

∫∞−∞ yψ(y)dy = 0. Let f : R → R be any

convex function. Then, if

a =n

∫∞−∞ ψ(y)dy2ψ(0)

,

one has ∫ ∞−∞

ψ(y)f(y)dy ≤ ψ(0)∫ a−a

(1 − |y|a

)n−1f(y)dy.



For the right-hand inclusion write

|Kx| =∫ 0−hK0 (−u)

φ(y)(1− λy)n+1 dy +

∫ hK0 (u)0

φ(y)(1 − λy)n+1 dy

≥∫ 0−hK0 (−u)

φ(y)(1 + (n+ 1)λy)dy

+∫ hK0(u)

0

φ(y)(1 + (n+ 1)λy +(n+ 1)(n+ 2)

2λ2y2)dy

= |K0|+ (n+ 1)(n+ 2)2

λ2∫ hK0 (u)

0

y2φ(y)dy,

where for the last equality we have used the fact that the center of gravity is at 0.Let a be such that∫ hK0 (u)

0

yφ(y)dy = φ(0)a2

n(n+ 1)=

∫ an

−ayψ0(y)dy,

where ψ0(y) = φ(0)(1 + ya )n−1.

Now one shows as in the beginning of the proof of Theorem 6 that for all functionsψ : R+0 → R+0 such that ψ

1n−1 is continuous on its support and continuous from

the right at 0, concave on its support and such that

(i) ψ(0) = φ(0),

(ii)∫ ∞

0

yψ(y)dy =∫ ∞

0

yφ(y)dy,

the integral∫∞0y2ψ(y)dy is minimal if ψ is of the form

ψ0(y) ={φ(0)(1 + ya )

n−1 if y ∈ [0, an ],0 otherwise,

and that for all functions ψ : R−0 → R+0 such that ψ1

n−1 is continuous on its supportand continuous from the left at 0, concave on its support and such that

(i) ψ(0) = φ(0),

(ii)∫ 0−∞

yψ(y)dy =∫ 0−∞

yφ(y)dy,

the integral∫ 0−∞ y

2ψ(y)dy is maximal if ψ is of the form

ψ0(y) ={φ(0)(1 + ya )

n−1 if y ∈ [−a, 0],0 otherwise.

Therefore∫ hK0 (u)0

y2φ(y)dy ≥∫ a

n

0

y2ψ0(y)dy =(1 + 1n )

n+1 − 2n(n+ 1)(n+ 2)

a3φ(0)

and ∫ 0−hK0(−u)

y2φ(y)dy ≤∫ 0−ay2ψ0(y)dy =

2n(n+ 1)(n+ 2)

a3φ(0).

Thus we get ∫ hK0 (u)0

y2φ(y)dy ≥ e− 22

∫ 0−hK0 (−u)

y2φ(y)dy,



and therefore ∫ hK0 (u)0

y2φ(y)dy ≥ e− 24

∫ hK0 (u)−hK0 (−u)

y2φ(y)dy.

It follows that

|Kx| ≥ |K0|(1 + (n+ 1)(n+ 2)(e− 2)8

λ2‖u‖2E(K0)),which implies, using (6),

S(K, t) ⊆ 2√

2((e− 2)(n+ 1)(n+ 2)) 12 (

tv2n|K||K0| )

12 (1− |K||K

0|tv2n

)12E(K0).

Next we estimate the “size” of S(K, t) in terms of the body K. We need thefollowing lemma.

Lemma 8 (see for instance [S2]). Let K be a convex body in Rn with center ofgravity at 0. Then

1e≤ 1|K|

∫ hK(u)0

φ(y)dy ≤ 1− 1e.

Theorem 9. Let K be a convex body in Rn. Then:(i)

(1− ( |K||K0|

tv2n)

1n )K ⊆ S(K, t) ⊆ (1− |K||K

0|etv2n

)K.

(ii) If in addition K is symmetric, then

(1 − ( |K||K0|

tv2n)

1n )K ⊆ S(K, t) ⊆ (1 − |K||K

0|tv2n

)12K.

Proof. Let u ∈ Sn−1, λ ∈ R, 0 ≤ λ ≤ 1, be given and let x = λhK0 (u)u. Then Kcontains αK + x for all α, 0 ≤ α ≤ 1− λ, and consequently Kx ⊆ 1αK0; thereforefor x ∈ ∂S(K, t) we have

t =|K||Kx|v2n

≤ |K||K0|

αnv2n,

and hence

α ≤ ( |K||K0|tv2n

)1n .

Thus for all λ with λ ≤ 1− (|K||K0|/tv2n)1n we have

λ

hK0(u)u ∈ S(K, t).

This proves the left-hand inclusion.For the right-hand inclusion we first treat the symmetric case.Let x = λu, u ∈ Sn−1, 0 ≤ λ < ||u||−1K . Let fλ be as in the proof of Theorem 6.

By (2) and symmetry,

|Kx| =∫ ||u||K

0

φ(y)fλ(y)dy.



Notice that for all functions ψ : R+0 → R+0 such that ψ1

n−1 is continuous on itssupport and continuous at 0 from the right, decreasing and concave on its supportand such that

(i) ψ > 0 on [0, ||u||K), ψ = 0 on [||u||K ,∞),

(ii)∫ ||u||K

0

ψ(y)dy =∫ ||u||K

0

φ(y)dy =|K0|

2,

the integral∫ ||u||K0 ψ(y)fλ(y)dy is smallest if ψ is of the form

ψ(y) ={c(1− y||u||K )n−1 if y ∈ [0, ||u||K),0 otherwise,

where

c =n|K0|2||u||K .

Hence

|Kx| ≥ c∫ ||u||K

0

((1− y||u||K )n−1(1− λy)n+1 +

(1 − y||u||K )n−1(1 + λy)n+1

)dy

=|K0|

1− λ2||u||2K,

which implies

S(K, t) ⊆ (1 − |K||K0|tv2n

)12K.

Next we consider the non-symmetric case:

|Kx| =∫ hK0 (u)

0

φ(y)(1− λy)n+1 dy +

∫ hK0 (−u)0

φ(−y)(1 + λy)n+1

dy

≥∫ hK0(u)

0

φ(y)(1− λy)n+1 dy.

Fix λ, and note again that among all functions ψ : R+0 → R+0 such that ψ1

n−1

is continuous on its support and continuous from the right at 0, decreasing andconcave on its support and for which

ψ > 0 on [0, hK0(u)), ψ = 0 on [hK0(u),∞),∫ hK0 (u)0

ψ(y)dy =∫ hK0 (u)

0

φ(y)dy,

the integral∫ hK0 (u)0

ψ(y)(1−λy)n+1dy is smallest if ψ is of the form

ψ(y) ={c(1− yhK0 (u) )

n−1 if y ∈ [0, hK0(u)),0 otherwise,

where

c =n

∫ hK0(u)0 φ(y)dyhK0(u)

.

Arguments similar to the previous ones, together with Lemma 8, then finish theproof.



II. Santaló-regions and affine surface area

Recall that for a convex body K in Rn the affine surface area is

as(K) =∫∂K

κ(x)1

n+1dµ(x),

where κ(x) is the (generalized) Gaussian curvature in x ∈ ∂K and µ is the surfacemeasure on ∂K. We prove here

Theorem 10. Let K be a convex body in Rn. Then

limt→∞ t

2n+1 (|K| − |S(K, t)|) = 1

2(|K|vn

)2

n+1as(K).

In the proof of Theorem 10 we follow the ideas of [S-W]. We need several lemmasfor the proof. We also use the following notations. For x ∈ ∂K, N(x) is the outerunit normal vector to ∂K in x. For two points x and y in Rn, [x, y] = {αx+(1−α)y :0 ≤ α ≤ 1} denotes the line segment from x to y.

The proof of the following lemma is standard.

Lemma 11. Let K and L be two convex bodies in Rn such that 0 ∈ int(L) andL ⊆ K. Then

|K| − |L| = 1n

∫∂K

〈x,N(x)〉(1 − ( ||xL||||x|| )n)dµ(x),

where xL = [0, x] ∩ ∂L.For x ∈ ∂K denote by r(x) the radius of the biggest Euclidean ball contained in

K that touches ∂K at x. More precisely,

r(x) = max{r : x ∈ B(y, r) ⊂ K for some y ∈ K}.Remark. It was shown in [S-W] that

(i) if B(0, 1) ⊂ K, thenµ{x ∈ ∂K : r(x) ≥ β} ≥ (1 − β)n−1voln−1(∂K),

(ii) ∫∂K

r(x)−αdµ(x) 0 andfor all t such that (S(K, t) has non-empty interior, we have

0 ≤ 1n〈x,N(x)〉t 2n+1

(1− (‖xt‖‖x‖ )

n

)≤ cr(x)− n−1n+1 ,

where xt = [0, x] ∩ ∂S(K, t) and c is a constant independent of x and t.Lemma 13. Suppose 0 is in the interior of K. Then the limit

limt→∞

1n〈x,N(x)〉t 2n+1

(1− (‖xt‖‖x‖ )

n

)



exists a.e. and is equal to12(|K|vn

)2

n+1κ(x)1

n+1 ,

where κ(x) is the Gaussian curvature at x ∈ ∂K.Proof of Theorem 10. We may assume that 0 is in the interior of K. By Lemma 11and with the notation of Lemma 12 we have

|K| − |S(K, t)| = 1n

∫∂K

〈x,N(x)〉(1 − ( ||xt||||x|| )n)dµ(x).

By Lemma 12 and the remark preceding it, the functions under the integral signare bounded uniformly in t by an L1-function, and by Lemma 13 they convergepointwise a.e. We apply Lebesgue’s convergence theorem.

Proof of Lemma 12. Let x ∈ ∂K be such that r(x) > 0. As ‖xt‖ = ‖x‖− ‖x− xt‖,we have

1n〈x,N(x)〉

(1− (‖xt‖‖x‖ )

n

)≤ 〈 x‖x‖ , N(x)〉‖x − xt‖.(7)

a) We consider first the case where

‖x− xt‖ < r(x)〈 x‖x‖ , N(x)〉.

Let ρ̃ = ‖xt− (x− r(x)N(x))‖. By assumption 0 < ρ̃ < r(x). Computing ρ̃, we getρ̃ = (||x− xt||2 + r(x)2 − 2r(x)||x − xt||〈 x‖x‖ , N(x)〉)

1/2.

Since K contains the Euclidean ball of radius r(x) centered at x−r(x)N(x), Kxt iscontained in the polar (with respect to xt) of the Euclidean ball with radius r(x).Hence by Corollary 5,

t =|K||Kxt|

v2n≤ |K|vnr(x)n(1− ( ρ̃r(x))2)

n+12

,

and therefore, using (7),

1n〈x,N(x)〉t 2n+1 (1− (‖xt‖‖x‖ )

n) ≤ ( |K|vn

)2

n+1 r(x)−n−1n+1 ,

which proves Lemma 12 in this case.b) Now we consider the case where

‖x− xt‖ ≥ r(x)〈 x‖x‖ , N(x)〉.

We can suppose that t is big enough so that xt 6= 0. We choose α > 0 such thatB(0, α) ⊆ K ⊆ B(0, 1α ) and t so big that xt /∈ B(0, α). K contains the sphericalcone C = co[x,H ∩ B(0, α)], where H is the hyperplane through 0 orthogonal tothe line segment [0, x]. We get

|Cxt | = vn−1||x||n

nαn−1||xt||(||x|| − ||xt||)n .

Consequently

t =|K||Kxt |

v2n≤ |K|vn−1‖x‖

n

nv2nαn−1‖xt‖ ‖x− xt‖n



and hence, using (7),

1n〈x,N(x)〉t 2n+1 (1− (‖xt‖‖x‖ )

n) ≤ ( |K|vn−1n v2n

)2

n+1 r(x)−n−1n+1

1

α4n

n+1.

Proof of Lemma 13. As in the proof of Lemma 12, we can choose an α > 0 suchthat

B(0, α) ⊆ K ⊆ B(0, 1α

).

Therefore

1 ≥ 〈 x‖x‖ , N(x)〉 ≥ α2.(8)

Since x and xt are collinear,

||x|| = ||xt||+ ||x− xt||,and hence

1n〈x,N(x)〉

(1− (‖xt‖‖x‖ )

n

)=

1n〈x,N(x)〉

((1− (1− ‖x− xt‖‖x‖ )

n

)≥ 〈 x‖x‖ , N(x)〉‖x − xt‖

(1− d · ‖x− xt‖‖x‖

),

(9)

for some constant d, if we choose t sufficiently large. We denote by θ the anglebetween x and N(x). Then 〈x/‖x‖, N(x)〉 = cosθ.

By [S-W], r(x) > 0 a.e., and by [L2] the Dupin indicatrix exists a.e. and is anelliptic cylinder or an ellipsoid.

(i) Case where the indicatrix is an ellipsoid. This case can be reduced to thecase of a sphere by an affine transformation with determinant 1 (see for instance[S-W]). Let

√ρ(x) be the radius of this sphere. Recall that we have to show that

limt→∞

1n〈x,N(x)〉t 2n+1

(1− (‖xt‖‖x‖ )

n

)=

12(|K|vn

)2

n+1ρ(x)−n−1n+1 .

We put ρ(x) = ρ and we introduce a coordinate system such that x = 0 and N(x) =(0, . . . 0,−1). H0 is the tangent hyperplane to ∂K at x = 0, and {Hs : s ≥ 0} isthe family of hyperplanes parallel to H0 that have non-empty intersection with Kand are at distance s from H0. For s > 0, H+s is the halfspace generated by Hsthat contains x = 0. For a ∈ R, let za = (0, . . . , 0, a), and let Ba = B(za, a) be theEuclidean ball with center za and radius a. As in [W], for ε > 0 we can choose s0so small that for all s ≤ s0

Bρ−ε ∩H+s ⊆ K ∩H+s ⊆ Bρ+ε ∩H+s .For λ ∈ R let Gλ = {x : 〈x, zρ+ε − xt〉 = λ} be a hyperplane orthogonal to theline segment [xt, zρ+ε], if t is sufficiently large. Let λ0 = max{λ : G+λ ∩ Bρ+ε ⊆H+s0 ∩Bρ+ε}. Define C to be the cone tangent to Bρ+ε at Gλ0 ∩Bρ+ε, and choosethe minimal λ1 so that

K ∩ {x : λ0 ≤ 〈x, zρ+ε − xt〉 ≤ λ1} ⊆ D = C ∩ {x : λ0 ≤ 〈x, zρ+ε − xt〉 ≤ λ1}.Then K is contained in the union of the truncated cone D of height h = |λ1 − λ0|and the cap L = {x ∈ Bρ+ε : 〈x, zρ+ε − xt〉 ≤ λ0} (see Figure 1).



ε+ρ

x

x t

s 0H 1

h 1

Figure 1. The estimate from below

ThereforeKxt ⊇ (D ∪ L)xt ,

and to estimate |Kxt | we have to compute |(D ∪ L)xt |. To do so we prove thefollowing more general result.

Claim 1. Let M be the convex body that is the union of a truncated spherical cone Dwith height h and a cap L of a Euclidean ball with radius r such that D is “tangent”to L. For a point x in L and on the axis of symmetry of M let a=distance(x,D),b=distance(x,∂L) and b0 =

rb+(a+b)(r−b)r−(a+b) (see Figure 2). Then, if x is such that

r > a+ b,

|Mx| = vn−1( 1rn

∫ 1r

(r−b)+b0

(1− y2)n−12(1− (r−b)yr )n+1

dy

+1n

(1b0

+1

a+ h)(

(2r(a + b)− (a+ b)2) 12rb+ (a+ b)(r − b) )

n−1).

Proof of Claim 1. We introduce a coordinate system such that x = 0 and the x1-axis coincides with the axis of symmetry of M (see Figure 2).

Notice now that M0 is such that each (n− 1)-dimensional section orthogonal tothe x1-axis is an (n− 1)-dimensional Euclidean ball with radius l(x1), where

l(x1) =((a+ h)x1 + 1)(2r(a+ b)− (a+ b)2) 12

(a+ h+ b0)(r − (a+ b)) , if −1

a+ h≤ x1 ≤ 1

b0,

l(x1) =1r((1 + x1(r − b))2 − r2x21)

12 , if

1b0≤ x1 ≤ 1

b.

From this Claim 1 follows.

Now we apply Claim 1 to our situation. Then

r = ρ+ ε and b = ρ+ ε− c,where

c2 = ||x− xt||2 + (ρ+ ε)2 − 2(ρ+ ε)||x − xt|| cos θ,



x=O

ah

b

x 1

Figure 2. Claim 1

a = c− ρ+ ε− s0c

(c2 − ||x− xt||2 sin2 θ) 12 − (2(ρ+ ε)s0 − s20)

12 ||x − xt|| sin θ

c,

and

b0 =(ρ+ ε)2

c− a − c.Therefore

|Kxt | ≥ vn−1(ρ+ ε)n

∫ 10

(1− y2)n−12(1 − cyρ+ε)n+1

dy

+1n

(1b0

+1

a+ h)vn−1rn−1D

− vn−1(ρ+ ε)n

∫ ρ+εb0+c

0

(1− y2)n−12(1 − cyρ+ε )n+1

dy,

where rD is the radius of the base of the spherical cone in (D ∪ L)xt . We put

R =1n

(1b0

+1

a+ h)vn−1rn−1D

− vn−1(ρ+ ε)n

∫ ρ+εb0+c

0

(1− y2)n−12(1− cyρ+ε )n+1

dy.

Then, by Lemma 4 (ii), for ε > 0

|Kxt | ≥ (1 − ε)vn2

n+12 (ρ+ ε)n( cρ+ε )

n+12 (1− cρ+ε )

n+12

+R,

provided thatc

ρ+ ε>

11 + 2ε3n

.

We choose t so big that this holds.



Hence

t =|K||Kxt |

v2n≥

(12

)n+12

((1− ε)|K|

vn

)(ρ+ ε)−n

( cρ+ε(1 − cρ+ε))n+1

2

×{

1 +2

n+12 R(ρ+ ε)n( cρ+ε )

n+12 (1 − cρ+ε)

n+12

(1− ε)vn

}and by (9), for some constant d,

t2

n+1

n〈x,N(x)〉

(1− (‖xt‖‖x‖ )

n

)

≥ 12((1 − ε)|K|

vn)

2n+1 (ρ+ ε)−

n−1n+1

(1 + k(2 ‖x−xt‖ cos θρ+ε − ‖x−xt‖2

(ρ+ε)2 ))−1(1 − d · ‖x−xt‖‖x‖ )

(1 + ‖x−xt‖2

(ρ+ε)2 − 2 ‖x−xt‖ cos θρ+ε )12 (1 − ‖x−xt‖2(ρ+ε) cos θ )

×{1 + 2n+12 R(ρ+ ε)n

(1− ε)vn (1−c

ρ+ ε)

n+12 (

c

ρ+ ε)

n+12 } 2n+1 ,

as

1− cρ+ ε

≤ ‖x− xt‖ cos θρ+ ε

(1− ‖x− xt‖2(ρ+ ε) cos θ

)(1 + k(2‖xxt‖ cos θρ+ ε

− ‖x− xt‖2

(ρ+ ε)2)),

for some constant k, if t is big enough. R remains bounded for t→∞.Note also that cos θ remains bounded from below by (8).Thus we have a lower bound for the expression in question.To get an upper bound we proceed in a similar way. For λ ∈ R, now let

Gλ = {x : 〈x, zρ−ε − xt〉 = λ} be a hyperplane orthogonal to the line segment[xt, zρ−ε]. Let λ0 = max{λ : G+λ ∩ Bρε ⊆ H+s0 ∩ Bρ−ε}. Let P be the point wherethe half-line starting at xt through zρ−ε intersects ∂K.

Let C be the spherical cone C = co[P,Bρ−ε ∩ Gλ0 ]. Let h be the height of thiscone (see Figure 3).

Let L = {x ∈ Bρ−ε : 〈x, zρ−ε − xt〉 ≤ λ0}. Then K ⊇ C ∪ L, and henceKxt ⊆ (C ∪ L)xt ,

and to estimate |Kxt | we have to compute (C ∪ L)xt .

To do so we prove the more general

Claim 2. Let M be the union of a spherical cone C with height h and a cap L ofa Euclidean ball with radius r such that the base of C coincides with the base ofL. For a point x in L and on the axis of symmetry of M let α = distance(x,C),β = distance(x, ∂L) (see Figure 4). Then, with β0 =

rβ+(α+β)(r−β)r(α+β) and x chosen

such that r > α+ β,

|Mx| = vn−1( 1rn

∫ 1r

(r−β)+β0

(1 − y2)n−12(1− (r−β)yr )n+1

dy

+1

nα(2r(α + β)− (α+ β)2)n12 ((α + β0)n

βn0− h

n

(α+ h)n)).



ε−ρ

P

x

x t

s 0

H 2

h 2

Figure 3. The estimate from above

x=O

ah b

x 1

Figure 4. Claim 2



Proof of Claim 2. We introduce a coordinate system such that x = 0 and the x1-axis coincides with the axis of symmetry of M (see Figure 4).

Notice now that M0 is such that each (n− 1)-dimensional section orthogonal tothe x1-axis is an (n− 1)-dimensional Euclidean ball with radius l(x1), where

l(x1) =αx1 + 1

(2r(α+ β)− (α + β)2) 12 , if −1

α+ h≤ x1 ≤ 1

β0,

l(x1) =1r((1 + x1(r − β))2 − r2x21)

12 , if

1β0

≤ x1 ≤ 1β.

From this Claim 2 follows.Now we apply Claim 2 to our situation. There

r = ρ− ε, β = ρ− ε− γ,where

γ2 = ||x− xt||2 + (ρ− ε)2 − 2(ρ− ε)||x− xt|| cos θ,

α = γ − ρ− ε− s0γ

(γ2 − ||x− xt||2 sin2 θ) 12 − (2(ρ− ε)s0 − s20)

12 ||x− xt|| sin θ

γ,

and

β0 =(ρ− ε)2γ − α − γ.

Then we get, similarily as before,

t2

n+1

n〈x,N(x)〉

(1− (‖xt‖‖x‖ )

n

)

≤ 12(|K|vn

)2

n+1 (ρ− ε)−n−1n+1 1(1 + ‖x−xt‖

2

(ρ−ε)2 − 2 ‖x−xt‖ cos θρ−ε )12 (1− ‖x−xt‖2(ρ−ε) cos θ )

×{1 + 2n+12 R(ρ− ε)n

vn(1− γ

ρ− ε )n+1

2 (γ

ρ− ε )n+1

2 } 2n+1 ,with a suitably defined R.

This finishes the proof of Lemma 13 in the case where the indicatrix is an ellip-soid.

(ii) Case where the Dupin indicatix is an elliptic cylinder. Recall that then wehave to show that

limt→∞

1n〈x,N(x)〉t 2n+1

(1− (‖xt‖‖x‖ )

n

)= 0.

We can again assume (see [S-W]) that the indicatrix is a spherical cylinder, i.e. theproduct of a k-dimensional plane and an (n− k− 1)-dimensional Euclidean sphereof radius ρ. Moreover we can assume that ρ is arbitrarily large (see also [S-W]).

By Lemma 9 of [S-W] we then have, for sufficiently small s and some ε > 0,

Bρ−ε ∩H+s ⊆ K ∩H+s .Using similar methods, this implies Lemma 13.



Proposition 14. Let K be a convex body such that ∂K is C3 and has strictlypositive Gaussian curvature everywhere. Then there is δ0 > 0 such that for allδ < δ0

S(K,vn−1

2(n+ 1)vnδ) ⊆ Kδ|K|.

Proof. As in the proof of Lemma 12, we can choose 1 > α > 0 such that

B(0, α) ⊆ K ⊆ B(0, 1α

).

Therefore we have, for all x ∈ ∂K,1 ≥ 〈 x‖x‖ , N(x)〉 ≥ α

2.

Let R0 = minx∈∂K,1≤i≤n−1Ri(x), where Ri(x) is the i-th principal radius of cur-vature at x ∈ ∂K. We know that R0 > 0 (see [L2]).

Let 1 > ε > 0 be given such that ε < min{R02 , 6nα4} and

(1− ε)(1− εR0

)n−1(1− ε6nα4

)n >12.

By assumption the Dupin indicatrix exists for all x ∈ ∂K and is an ellipsoid. Forx ∈ ∂K given, we can assume that, after an affine transformation, the indicatrix atx is a Euclidean sphere. Let

√ρ(x) be the radius of this Euclidean sphere. Note

that, for all x ∈ ∂K,ρ(x) ≥ R0.(10)

Then, with the notation used in the proof of Lemma 13, there exists s(x) > 0 suchthat

Bρ(x)−ε ∩H+s(x) ⊆ K ∩H+s(x) ⊆ Bρ(x)+ε ∩H+s(x).Let s1 = minx∈∂K s(x). We know that s1 > 0, as ∂K is C3 and compact. Let

s0 = min{s1, (R0 − ε)(1− 11 + 2ε3n)}.(11)

Let δ0 > 0 be so small that for all x ∈ ∂K two conditions are satisfied: first||x− xδ0 ||〈

x

||x|| , N(x)〉 ≤s02,(12)

where xδ0 = [0, x] ∩ ∂Kδ0|K|, and, second,H+δ0 ∩Bρ(x)−ε ⊆ H+s0 ∩Bρ(x)−ε,

where Hδ0 is the hyperplane through xδ0 that cuts off exactly δ0|K| from K.Suppose now that the above proposition is not true. Then there is δ < δ0 and

xs ∈ ∂S(K, vn−12(n+1)vnδ ) such that xs /∈ Kδ|K|. Let x ∈ ∂K be such that xs ∈ [0, x].We also can assume that the Dupin indicatrix at x is a Euclidean ball with radius√ρ(x). We choose xδ ∈ ∂Kδ|K| such that xδ ∈ [0, x]. Then

||x− xδ||〈 x||x|| , N(x)〉 < ||x− xs||〈x

||x|| , N(x)〉.

By constructionδ|K| = |K ∩H+δ | ≥ |Bρ(x)−ε ∩H+δ |

≥ minH∈H

|Bρ(x)−ε ∩H+| = |Bρ(x)−ε ∩H+0 |,



where

H = {H : H is a hyperplane through xδ, x ∈ H+}.For the height h of this cap |Bρ(x)−ε ∩H+0 | of Bρ(x)−ε of minimal volume we have

h ≥ ||x− xδ||〈 x||x|| , N(x)〉(1 −||x− xδ||

2〈 x||x|| , N(x)〉(ρ(x) − ε)).

Using (10), (11) and (12), we get

h ≥ ||x− xδ||〈 x||x|| , N(x)〉(1 −ε

6nα4).

The volume of a cap of a Euclidean ball with radius r and height h can beestimated from below by

≥ vn−12n+12 r

n−12 h

n+12

n+ 1(1 − h

2r)

n+12 .

Therefore

δ|K| ≥ vn−1n+ 1

2n+12 ρ(x)

n−12 (||x − xδ||〈 x||x|| , N(x)〉)

n+12 (1 − ε

R0)

n−12 (1− ε

6nα4)n,

where we have used again (10), (11), (12) and the fact that 〈 x||x|| , N(x)〉 ≥ α2. Thus

(||x− xs||〈 x||x|| , N(x)〉)n+1

2 <(n+ 1)δ|K|ρ(x)−n−12

vn−12n+12 (1− ε6nα4 )n(1 − εR0 )

n−12

.(13)

Since, in the notation of Lemma 13 and using (10), (11) and (12),

c

ρ(x) + ε≥ 1−

||x− xs||〈 x||x|| , N(x)〉ρ(x) + ε

>1

1 + 2ε3n,

the estimate from below from Lemma 13 for |K||Kx|/v2n holds for x = xs, and weget (see p. 17)

|K||Kxs|v2n

>(1− ε)|K|ρ(x)−n(1 + ερ(x) )−n

2n+12 vn( cρ(x)+ε )

n+12 (1− cρ(x)+ε )

n+12

.

Now notice thatc

ρ(x) + ε≤ 1

and

1− cρ(x) + ε

≤||x− xs||〈 x||x|| , N(x)〉

ρ(x)(1 + ερ(x).

Therefore (13) implies that

|K||Kxs |v2n

> (1− ε)(1− εR0

)n−1(1− ε6nα4

)nvn−1

(n+ 1)δvn.

This is a contradiction.



References

[B] K. Ball: Logarithmically concave functions and sections of convex sets in Rn, StudiaMath. vol. 88 (1988),69-84. MR 89e:52002

[D-H] G. Dolzmann, D. Hug: Equality of two representations of extended affine surface area,Archiv der Mathematik 65 no.4 (1995),352-356. MR 97c:52019

[F] M. Fradelizi: Hyperplane sections of convex bodies in isotropic position, to appear inContributions to Algebra and Geometry.

[G] P. Gruber: Aspects of approximation of convex bodies, Handbook of Convex Geometry,vol.A(1993), 321-345, North Holland. MR 95b:52003

[He] D. Hensley: Slicing convex bodies—bounds for slice area in terms of the body’s covari-ance, Proc. Amer. Math. Soc. 79(4) (1980), 619-625. MR 81j:52008

[L1] K. Leichtweiss: Zur Affinoberfläche konvexer Körper, Manuscripta Mathematica 56(1986), 429-464. MR 87k:52011

[L2] K. Leichtweiss: Über eine Formel Blaschkes zur Affinoberfläche, Studia Scient. Math.Hung. 21 (1986), 453-474. MR 89b:53016

[Lu] E. Lutwak: Extended affine surface area, Adv. in Math. 85 (1991), 39-68. MR92d:52012

[Lu-O] E. Lutwak, V.Oliker: On the regularity of solutions to a generalization of the Minkowskiproblem, J. Differential Geometry, vol. 41 (1995), 227-246. MR 95m:52007

[Me-P] M. Meyer, A. Pajor: On the Blaschke Santaló inequality, Arch. Math. 55(1990), 82-93.MR 92b:52013

[Mi-P] V. D. Milman, A. Pajor: Isotropic position and inertia ellipsoids and zonoids of the unitball of a normed n-dimensional space, GAFA (eds. J. Lindenstrauss, V.D. Milman),Lecture Notes in Math.1376, Springer, Berlin (1989), 64-104. MR 90g:52003

[Sch] R. Schneider: Convex bodies: The Brunn-Minkowski theory, Encyclopedia of Math.and its Applic.44, Cambridge University Press (1993). MR 94d:52007

[S1] C. Schütt: On the affine surface area, Proc. Amer. Math. Soc. 118 (1993), 1213-1218.MR 93j:52009

[S2] C. Schütt: Floating body, illumination body and polytopal approximation, C. R. Acad.Sci. Paris Sér. I Math. 324 (1997), 201-203. MR 98b:52005

[S-W] C. Schütt, E. Werner: The convex floating body, Math. Scand. 66 (1990), 275-290. MR91i:52005

[W] E. Werner: Illumination bodies and affine surface area, Studia Math. 110 (1994), 257-269. MR 95g:52010

Université de Marne-La-Valee, Equipe d’Analyse et de Mathématiques Appliquees,Cité Descartes-5, bd Descartes-Champs-sur-Marne, 77454 Marne-la-Vallée cedex 2,France

E-mail address: [email protected]

Department of Mathematics, Case Western Reserve University, Cleveland, Ohio44106

E-mail address: [email protected] address: Université de Lille 1, UFR de Mathématiques, 59655 Villeneuve d’Ascq,

France


Documents

THE SANTALO-REGIONS OF A CONVEX BODY · 2018. 11. 16. · Abstract. Motivated by the Blaschke-Santal o inequality, we de ne for a con-vex body K in Rn and for t 2 R the Santal o-regions